Since no current flows through AD. so the given circuit is a balanced Wheatstone bridge.
∴ The condition of balanced Wheatstone bridge is
\(\frac{P}{Q} = \frac{R}{S}\)
or \(\frac{2}{4} = \frac{3}{x}\)
or X = 6 Ω

In the circuit ABDC, the resistances P and Q are in series and R and X are in series and both are in parallel.
So equivalent resistance,

∴ Current drawn by the circuit,
I = \(\frac{E}{3.6+2.4}= \frac{6}{6}\) 1A